Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\end{matrix}\right.\)

PT ion rút gọn:        \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

Ban đầu:                0,01       0,015

Pư:                         0,01---->0,01

Sau pư:                   0           0,005        0,01

=> \(m_X=m_{BaSO_4}=0,01.233=2,33\left(g\right)\)

$\rm a)$
$\rm 2Mg + CO_2 \xrightarrow{t^o} 2MgO + C$
$\rm C + CO_2 \xrightarrow{t^o} 2CO$
$\rm CO + CuO \xrightarrow{t^o} Cu + CO_2$
$\rm CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O$
$\rm CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2$
$\rm Ca(HCO_3)_2 \xrightarrow{t^o} CaCO_3 \downarrow + CO_2 \uparrow + H_2O$
$\rm b)$
$\rm CO_2 + H_2O + Na_2CO_3 \rightarrow 2NaHCO_3$
$\rm NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$
$\rm Na_2CO_3 + Ca(HCO_3)_2 \rightarrow CaCO_3 \downarrow + 2NaHCO_3$
$\rm NaHCO_3 + HCl \rightarrow NaCl + CO_2 \uparrow + H_2O$

Ta có: \(\left\{{}\begin{matrix}m_{bình.đựng.H_2SO_{4\left(đặc\right)}.tăng}=m_{H_2O}=2,52\left(g\right)\\m_{CaCO_3}=m_{kết.tủa}=14\left(g\right)\end{matrix}\right.\)

Theo BTNT: \(\left\{{}\begin{matrix}n_H=2n_{H_2O}=2.\dfrac{2,52}{18}=0,28\left(mol\right)\\n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{14}{100}=0,14\left(mol\right)\\n_{O\left(X\right)}=\dfrac{3,0,8-0,28-0,14.12}{16}=0,07\left(mol\right)\end{matrix}\right.\)

=> \(n_C:n_H:n_O=0,14:0,28:0,07=2:4:1\)

=> CTPT của X có dạng \(\left(C_2H_4O\right)_n\)

Mặt khác: \(n_X=n_{N_2}=\dfrac{1,54}{28}=0,055\left(mol\right)\)

=> \(M_X=\dfrac{4,84}{0,055}=88\left(g/mol\right)\)

=> \(n=\dfrac{88}{44}=2\left(t/m\right)\)

=> CTPT của X là \(C_4H_8O_2\)

`CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O`

`CaCO_3 + CO_2 + H_2O -> Ca(HCO_3)_2`

`Ca(HCO_3)_2 + 2NaOH -> CaCO_3 + Na_2CO_3 + 2H_2O`

`CaCO_3 + CO_2 + H_2O -> Ca(HCO_3)_2`

`Ca(HCO_3)_2 + 2HCl -> CaCl_2 + 2CO_2 + 2H_2O`

`CaCl_2 + Na_2CO_3 -> CaCO_3 + 2NaCl`

1,12 lít khí thoát ra là N₂ => \(n_{N_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Có: \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

PTHH: \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                                0,3<------0,3

Ta có: \(m_{dd.giảm}=m_{CaCO_3}-m_{CO_2}-m_{H_2O}\)

=> 8,7 = 30 - 0,3.44 - m_H2O

=> \(m_{H_2O}=8,1\left(g\right)\)

=> \(n_{H_2O}=\dfrac{8,1}{18}=0,45\left(mol\right)\)

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,3\left(mol\right)\\n_H=2n_{H_2O}=0,9\left(mol\right)\\n_N=2n_{N_2}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(a=m_C+m_H+m_N=0,3.12+0,9+0,1.14=5,9\left(g\right)\)

=> Chọn A

Ta có: \(\left\{{}\begin{matrix}m_{bình.1.tăng}=m_{H_2O}=2,7\left(g\right)\\n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\end{matrix}\right.\)

Theo BTNT: \(\left\{{}\begin{matrix}n_H=2n_{H_2O}=2.\dfrac{2,7}{18}=0,3\left(mol\right)\\n_C=n_{Na_2CO_3}=0,2\left(mol\right)\end{matrix}\right.\)

=> \(n_{O\left(A\right)}=\dfrac{4,3-0,2.12-0,3}{16}=0,1\left(mol\right)\)

=> \(n_C:n_H:n_O=0,2:0,3:0,1=2:3:1\)

=> CTĐGN của A là C₂H₃O 

=> Chọn A

Câu 7:

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{1,76}{44}=0,04\left(mol\right)\\n_H=2.n_{H_2O}=2.\dfrac{1,26}{18}=0,14\left(mol\right)\\n_N=2n_{N_2}=2.\dfrac{0,224}{22,4}=0,02\left(mol\right)\\n_{O\left(Y\right)}=\dfrac{0,9-0,04.12-0,14-0,02.16}{16}=0\left(mol\right)\end{matrix}\right.\)

=> Y không chứa O

=> \(n_C:n_H:n_N=0,04:0,14:0,02=2:7:1\)

=> CTĐGN của Y là C₂H₇N

=> A

Câu 8:

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{1,8}{18}=0,2\left(mol\right)\\n_{O\left(M\right)}=\dfrac{3-0,1.12-0,2}{16}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(n_C:n_H:n_O=0,1:0,2:0,1=1:2:1\)

=> CTĐGN của M là CH₂O

=> Chọn C

Câu 2.

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol;n_{H_2O}=\dfrac{2,7}{18}=0,15mol\)

\(BTO:2n_{O_2}=2n_{CO_2}+n_{H_2O}=2\cdot0,1+0,15=0,35mol\)

\(n_{O_2}=\dfrac{0,35}{2}=0,175mol\)

\(\Rightarrow V_{O_2}=0,175\cdot22,4=3,92l\)

Chọn C.